Cornelsen Fokus Physik Lösungen

Your best friend weighs 81.5 kg and is a rugby player. In one of his games, he slides to a stop in a phenomenal manner. The coef

Alex777 [14]

A. The acceleration during the slide is 6.86 m/s²

B. The time taken to slide until he stops is 1.2 s

<h3>How to determine the force of friction</h3>

Mass (m) = 81.5 Kg
Coefficient of friction (μ) = 0.7
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
Frictional force (F) =?

F = μN

F = 0.7 × 798.7

F = 559.09 N

<h3>A. How to determine the acceleration</h3>

Mass (m) = 81.5 Kg
Frictional force (F) = 559.09 N
Acceleration (a) =?

a = F / m

a = 559.09 / 81.5

a = 6.86 m/s²

<h3>B. How to determine the time </h3>

Initial velocity (u) = 8.23 m/s
Final velocity (v) = 0 m/s
Decceleration (a) = -6.86 m/s²
Time (t) =?

a = (v – u) / t

t = (v – u) / a

t = (0 – 8.23) / -6.86

t = 1.2 s

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

4 0

1 year ago

A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simp

masha68 [24]

To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as

$T = 2\pi \sqrt{\frac{l}{g}}$

Where

l = Length

g = Gravity

Re-arrange to find L,

$L = g (\frac{T}{2\pi})^2$

Our values are given as

$g = 9.81m/s$

$T = 10.1s$

Replacing,

$L = g (\frac{T}{2\pi})^2$

$L = (9.81) (\frac{10.1}{2\pi})^2$

$L = 25.348m$

Therefore the height would be 25.348m

5 0

3 years ago

What timeis the name of the units that

Westkost [7]

Answer:

Joule is the name of the unit that are normally used to measure energy

Explanation:

7 0

1 year ago

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

2 years ago

What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze

yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of $62.5 cm of H_2O$

i.e. $h=62.5 cm =0.625 m$

Effective area $A=51 cm^2$

initial Pressure $= 1 atm=101.325 kPa$

Gauge Pressure $P=\rho gh$

$\rho =density\ of\ water =1000 kg/m^3$

$P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa$

Force creates a pressure of $P_1$ which will be equal to Gauge Pressure

$P_1=\frac{F}{A}$

$P_1=P_{gauge}$

$\frac{F}{A}=5.937 kPa$

$F=5.937\times 51\times 10^{-4}\times 10^3$

$F=30.27 N$

6 0

3 years ago